The focal lengths of the objective and eyepie | vedclass

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(B) Given: Focal length of objective $f_0 = 60\, cm$, focal length of eyepiece $f_e = 5\, cm$, and least distance of distinct vision $D = 25\, cm$.For

Vedclass · Accepted Answer

$14.4, 64.17\, cm$

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(B) Given: Focal length of objective $f_0 = 60\, cm$, focal length of eyepiece $f_e = 5\, cm$, and least distance of distinct vision $D = 25\, cm$.For an astronomical telescope, when the final image is formed at the least distance of distinct vision, the magnifying power $MP$ is given by:$MP = -\frac{f_0}{f_e} \left( 1 + \frac{f_e}{D} ight)$Substituting the values:$MP = -\frac{60}{5} \left( 1 + \frac{5}{25} ight) = -12 \left( 1 + 0.2 ight) = -12 	imes 1.2 = -14.4$(The magnitude of magnifying power is $14.4$)The length of the telescope $L$ is given by $L = f_0 + |u_e|$, where $u_e$ is the object distance for the eyepiece.Using the lens formula for the eyepiece: $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$Here, $v_e = -D = -25\, cm$ and $f_e = 5\, cm$.$\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25}$$|u_e| = \frac{25}{6} \approx 4.17\, cm$Therefore, $L = 60 + 4.17 = 64.17\, cm$.

The focal lengths of the objective and eyepiece of an astronomical telescope are $60\, cm$ and $5\, cm$ respectively. When the final image is formed at the least distance of distinct vision $(25\, cm)$, the magnifying power and the length of the telescope will be, respectively:

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